Question: What is the average value of $\dfrac{1}{x}$ on the interval $4\leq x \leq 8$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{3}{16}$ (Choice B) B $\dfrac{1}{16}$ (Choice C) C $\dfrac{\ln(32)}{4}$ (Choice D) D $\dfrac{\ln(2)}{4}$
Explanation: In general, this is the average value of function $f$ over the interval $[a,b]$ : $\dfrac{\int_a^b f(x)\,dx}{b-a}$ In our case, ${f(x)=\dfrac{1}{x}}$, ${a=4}$ and ${b=8}$ : $\begin{aligned} \dfrac{\int_{ a}^{ b} {f(x)}\,dx}{ b- a}&=\dfrac{\int_{{4}}^{ {8}} \left({\dfrac{1}{x}}\right)\,dx}{{8}-{4}} \\\\ &=\dfrac{\Big[\ln|x|\Big]_{4}^{8}}{4} \\\\ &=\dfrac{\ln|8|-\ln|4|}{4} \\\\ &=\dfrac{\ln(2)}{4} \end{aligned}$ In conclusion, the average value of $\dfrac{1}{x}$ on the interval $4\leq x \leq 8$ is $\dfrac{\ln(2)}{4}$.